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finlep [7]
2 years ago
5

How will you prepare 10mL of 0.050M HCl from a 0.10M solution of HCl

Chemistry
1 answer:
vichka [17]2 years ago
7 0
M₁V₁ = M₂V₂

(0.050M)*(10 mL) = (0.10M)*(x)

x = 5 mL

You would take 5 mL of the 0.10 M solution of HCl and add 5 mL of deionized water. Therefore, the total volume will be 10 mL and it will have a molarity of 0.050.
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For each solution, determine the p-values for each ion indicated. A solution that is 0.493 M in NaCl and 0.314 M in NH 4 Cl .
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Complete Question:

Ions to calculate the p-values: Na⁺, Cl⁻, and NH₄⁺

Answer:

pNa = 0.307

pCl = 0.093

pNH₄ = 0.503

Explanation:

The p-value is calculated by the antilog of the concentration of the substance of interest. For example, pH = -log[H⁺]. Thus, first, let's find the ions concentration.

Both substances are salts that solubilize completely, thus, by the solution reactions:

NaCl → Na⁺ + Cl⁻

NH₄Cl → NH₄⁺ + Cl⁻

So, for both reactions the stoichiometry is 1:1:1 and the concentration of the ions is equal to the concentration of the salts.

[Na⁺] = 0.493 M

[Cl⁻] = 0.493 + 0.314 = 0.807 M

[NH₄⁺] = 0.314 M

The p-values are:

pNa = -log[Na⁺] = -log(0.493) = 0.307

pCl = -log[Cl⁻] = -log(0.807) = 0.093

pNH₄ = -log[NH₄⁺] = -log(0.314) = 0.503

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